Mixing
on the inverse of trigonometric or/ and hyperbolic functions
$begingroup$
If we want to find, say, the inverse $tan$ function, $tan^{-1}$, in terms of (complex) logarithm function we start with the equation $z=tan w =frac{sin w}{cos w}=frac{1}{i}frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}}$, and solve it for $w$, which gives $e^{2iw}=frac{i-z}{i+z}$.
I have a doubt exactly at this stage: In the last equation if we take the logarithm of both sides, then we get $log (e^{2iw})=logleft(frac{i-z}{i+z}right)$. From this point on I saw in many books saying that
$$2iw=logleft(frac{i-z}{i+z}right)tag1$$
(and so that $w=frac {1}{2i}logleft(frac{i-z}{i+z}right)$). However in general we have that $log (e^{2iw})neq 2iw$. My question is that how do we get the equality $(1)$ from $log (e^{2iw})=logleft(frac{i-z}{i+z}right)$.
complex-analysis trigonometry logarithms inverse-function
edited Jan 19 at 21:11
José Carlos Santos
164k22132235
asked Jan 19 at 21:04
serenusserenus
24316
$endgroup$
add a comment |
$begingroup$
If we want to find, say, the inverse $tan$ function, $tan^{-1}$, in terms of (complex) logarithm function we start with the equation $z=tan w =frac{sin w}{cos w}=frac{1}{i}frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}}$, and solve it for $w$, which gives $e^{2iw}=frac{i-z}{i+z}$.
I have a doubt exactly at this stage: In the last equation if we take the logarithm of both sides, then we get $log (e^{2iw})=logleft(frac{i-z}{i+z}right)$. From this point on I saw in many books saying that
$$2iw=logleft(frac{i-z}{i+z}right)tag1$$
(and so that $w=frac {1}{2i}logleft(frac{i-z}{i+z}right)$). However in general we have that $log (e^{2iw})neq 2iw$. My question is that how do we get the equality $(1)$ from $log (e^{2iw})=logleft(frac{i-z}{i+z}right)$.
complex-analysis trigonometry logarithms inverse-function
edited Jan 19 at 21:11
José Carlos Santos
164k22132235
asked Jan 19 at 21:04
serenusserenus
24316
$endgroup$
add a comment |
$begingroup$
If we want to find, say, the inverse $tan$ function, $tan^{-1}$, in terms of (complex) logarithm function we start with the equation $z=tan w =frac{sin w}{cos w}=frac{1}{i}frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}}$, and solve it for $w$, which gives $e^{2iw}=frac{i-z}{i+z}$.
I have a doubt exactly at this stage: In the last equation if we take the logarithm of both sides, then we get $log (e^{2iw})=logleft(frac{i-z}{i+z}right)$. From this point on I saw in many books saying that
$$2iw=logleft(frac{i-z}{i+z}right)tag1$$
(and so that $w=frac {1}{2i}logleft(frac{i-z}{i+z}right)$). However in general we have that $log (e^{2iw})neq 2iw$. My question is that how do we get the equality $(1)$ from $log (e^{2iw})=logleft(frac{i-z}{i+z}right)$.
complex-analysis trigonometry logarithms inverse-function
edited Jan 19 at 21:11
José Carlos Santos
164k22132235
asked Jan 19 at 21:04
serenusserenus
24316
$endgroup$
If we want to find, say, the inverse $tan$ function, $tan^{-1}$, in terms of (complex) logarithm function we start with the equation $z=tan w =frac{sin w}{cos w}=frac{1}{i}frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}}$, and solve it for $w$, which gives $e^{2iw}=frac{i-z}{i+z}$.
I have a doubt exactly at this stage: In the last equation if we take the logarithm of both sides, then we get $log (e^{2iw})=logleft(frac{i-z}{i+z}right)$. From this point on I saw in many books saying that
$$2iw=logleft(frac{i-z}{i+z}right)tag1$$
(and so that $w=frac {1}{2i}logleft(frac{i-z}{i+z}right)$). However in general we have that $log (e^{2iw})neq 2iw$. My question is that how do we get the equality $(1)$ from $log (e^{2iw})=logleft(frac{i-z}{i+z}right)$.
complex-analysis trigonometry logarithms inverse-function
complex-analysis trigonometry logarithms inverse-function
edited Jan 19 at 21:11
José Carlos Santos
164k22132235
asked Jan 19 at 21:04
serenusserenus
24316
edited Jan 19 at 21:11
José Carlos Santos
164k22132235
asked Jan 19 at 21:04
serenusserenus
24316
edited Jan 19 at 21:11
José Carlos Santos
164k22132235
edited Jan 19 at 21:11
José Carlos Santos
164k22132235
edited Jan 19 at 21:11
José Carlos Santos
164k22132235
164k22132235
asked Jan 19 at 21:04
serenusserenus
24316
asked Jan 19 at 21:04
serenusserenus
24316
asked Jan 19 at 21:04
serenusserenus
24316
24316
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You must keep in mind that $log$ is not a function. In fact, every $zinmathbb{C}setminus{0}$ has infinitely many logarithms. To be more precise, if $w$ is a logarithm of $z$, then$${text{logarithms of }z}={w+2kpi i,|,kinmathbb Z}.$$
In particular $log e^z=log w$ simply means that $z$ itself is a logarithm of $w$.
answered Jan 19 at 21:08
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
$begingroup$
What do you mean when saying "some"? Could you please write your answer a bit more explicitly?
$endgroup$
– serenus
Jan 19 at 21:14
$begingroup$
I've edited my answer. What do you think now?
$endgroup$
– José Carlos Santos
Jan 19 at 21:24
1
$begingroup$
I think that $log (e^{2iw})={2iw+i2npi: nin mathbb{Z}}$, and since the sets $log(frac{i-z}{i+z})$ , $log(frac{i-z}{i+z})+i2npi$, are the same, as a consequence we get (1). Am I right?
$endgroup$
– serenus
Jan 19 at 21:42
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Jan 19 at 21:44
$begingroup$
Although my above statement is true, now I really see what you mean: Given a fixed $zinmathbb{C}$ , the set of solutions for the equation $e^w=z$ is ${ln|z|+i(Argz+2nπ):n inmathbb{Z}}$, which is denoted by $logz$, and although $w$ is a member of the set $logz$, notationally, we write $w=logz$. Now, substituting $w$ by $2iw$ and $z$ by $frac{i−z}{i+z}$, exactly in the same sense we get $2iwin log(frac{i−z}{i+z})$, or, by the usual notation, $2iw=log(frac{i−z}{i+z})$. What was misleading for me is the statement "taking logarithm of both sides". Thanks for the answer.
$endgroup$
– serenus
Jan 20 at 9:34
|
show 1 more comment
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You must keep in mind that $log$ is not a function. In fact, every $zinmathbb{C}setminus{0}$ has infinitely many logarithms. To be more precise, if $w$ is a logarithm of $z$, then$${text{logarithms of }z}={w+2kpi i,|,kinmathbb Z}.$$
In particular $log e^z=log w$ simply means that $z$ itself is a logarithm of $w$.
answered Jan 19 at 21:08
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
$begingroup$
What do you mean when saying "some"? Could you please write your answer a bit more explicitly?
$endgroup$
– serenus
Jan 19 at 21:14
$begingroup$
I've edited my answer. What do you think now?
$endgroup$
– José Carlos Santos
Jan 19 at 21:24
1
$begingroup$
I think that $log (e^{2iw})={2iw+i2npi: nin mathbb{Z}}$, and since the sets $log(frac{i-z}{i+z})$ , $log(frac{i-z}{i+z})+i2npi$, are the same, as a consequence we get (1). Am I right?
$endgroup$
– serenus
Jan 19 at 21:42
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Jan 19 at 21:44
$begingroup$
Although my above statement is true, now I really see what you mean: Given a fixed $zinmathbb{C}$ , the set of solutions for the equation $e^w=z$ is ${ln|z|+i(Argz+2nπ):n inmathbb{Z}}$, which is denoted by $logz$, and although $w$ is a member of the set $logz$, notationally, we write $w=logz$. Now, substituting $w$ by $2iw$ and $z$ by $frac{i−z}{i+z}$, exactly in the same sense we get $2iwin log(frac{i−z}{i+z})$, or, by the usual notation, $2iw=log(frac{i−z}{i+z})$. What was misleading for me is the statement "taking logarithm of both sides". Thanks for the answer.
$endgroup$
– serenus
Jan 20 at 9:34
|
show 1 more comment
$begingroup$
You must keep in mind that $log$ is not a function. In fact, every $zinmathbb{C}setminus{0}$ has infinitely many logarithms. To be more precise, if $w$ is a logarithm of $z$, then$${text{logarithms of }z}={w+2kpi i,|,kinmathbb Z}.$$
In particular $log e^z=log w$ simply means that $z$ itself is a logarithm of $w$.
answered Jan 19 at 21:08
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
$begingroup$
What do you mean when saying "some"? Could you please write your answer a bit more explicitly?
$endgroup$
– serenus
Jan 19 at 21:14
$begingroup$
I've edited my answer. What do you think now?
$endgroup$
– José Carlos Santos
Jan 19 at 21:24
1
$begingroup$
I think that $log (e^{2iw})={2iw+i2npi: nin mathbb{Z}}$, and since the sets $log(frac{i-z}{i+z})$ , $log(frac{i-z}{i+z})+i2npi$, are the same, as a consequence we get (1). Am I right?
$endgroup$
– serenus
Jan 19 at 21:42
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Jan 19 at 21:44
$begingroup$
Although my above statement is true, now I really see what you mean: Given a fixed $zinmathbb{C}$ , the set of solutions for the equation $e^w=z$ is ${ln|z|+i(Argz+2nπ):n inmathbb{Z}}$, which is denoted by $logz$, and although $w$ is a member of the set $logz$, notationally, we write $w=logz$. Now, substituting $w$ by $2iw$ and $z$ by $frac{i−z}{i+z}$, exactly in the same sense we get $2iwin log(frac{i−z}{i+z})$, or, by the usual notation, $2iw=log(frac{i−z}{i+z})$. What was misleading for me is the statement "taking logarithm of both sides". Thanks for the answer.
$endgroup$
– serenus
Jan 20 at 9:34
|
show 1 more comment
$begingroup$
You must keep in mind that $log$ is not a function. In fact, every $zinmathbb{C}setminus{0}$ has infinitely many logarithms. To be more precise, if $w$ is a logarithm of $z$, then$${text{logarithms of }z}={w+2kpi i,|,kinmathbb Z}.$$
In particular $log e^z=log w$ simply means that $z$ itself is a logarithm of $w$.
answered Jan 19 at 21:08
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
You must keep in mind that $log$ is not a function. In fact, every $zinmathbb{C}setminus{0}$ has infinitely many logarithms. To be more precise, if $w$ is a logarithm of $z$, then$${text{logarithms of }z}={w+2kpi i,|,kinmathbb Z}.$$
In particular $log e^z=log w$ simply means that $z$ itself is a logarithm of $w$.
answered Jan 19 at 21:08
José Carlos SantosJosé Carlos Santos
164k22132235
edited Jan 19 at 21:23
answered Jan 19 at 21:08
José Carlos SantosJosé Carlos Santos
164k22132235
answered Jan 19 at 21:08
José Carlos SantosJosé Carlos Santos
164k22132235
answered Jan 19 at 21:08
José Carlos SantosJosé Carlos Santos
164k22132235
164k22132235
$begingroup$
What do you mean when saying "some"? Could you please write your answer a bit more explicitly?
$endgroup$
– serenus
Jan 19 at 21:14
$begingroup$
I've edited my answer. What do you think now?
$endgroup$
– José Carlos Santos
Jan 19 at 21:24
1
$begingroup$
I think that $log (e^{2iw})={2iw+i2npi: nin mathbb{Z}}$, and since the sets $log(frac{i-z}{i+z})$ , $log(frac{i-z}{i+z})+i2npi$, are the same, as a consequence we get (1). Am I right?
$endgroup$
– serenus
Jan 19 at 21:42
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Jan 19 at 21:44
$begingroup$
Although my above statement is true, now I really see what you mean: Given a fixed $zinmathbb{C}$ , the set of solutions for the equation $e^w=z$ is ${ln|z|+i(Argz+2nπ):n inmathbb{Z}}$, which is denoted by $logz$, and although $w$ is a member of the set $logz$, notationally, we write $w=logz$. Now, substituting $w$ by $2iw$ and $z$ by $frac{i−z}{i+z}$, exactly in the same sense we get $2iwin log(frac{i−z}{i+z})$, or, by the usual notation, $2iw=log(frac{i−z}{i+z})$. What was misleading for me is the statement "taking logarithm of both sides". Thanks for the answer.
$endgroup$
– serenus
Jan 20 at 9:34
|
show 1 more comment
$begingroup$
What do you mean when saying "some"? Could you please write your answer a bit more explicitly?
$endgroup$
– serenus
Jan 19 at 21:14
$begingroup$
I've edited my answer. What do you think now?
$endgroup$
– José Carlos Santos
Jan 19 at 21:24
1
$begingroup$
I think that $log (e^{2iw})={2iw+i2npi: nin mathbb{Z}}$, and since the sets $log(frac{i-z}{i+z})$ , $log(frac{i-z}{i+z})+i2npi$, are the same, as a consequence we get (1). Am I right?
$endgroup$
– serenus
Jan 19 at 21:42
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Jan 19 at 21:44
$begingroup$
Although my above statement is true, now I really see what you mean: Given a fixed $zinmathbb{C}$ , the set of solutions for the equation $e^w=z$ is ${ln|z|+i(Argz+2nπ):n inmathbb{Z}}$, which is denoted by $logz$, and although $w$ is a member of the set $logz$, notationally, we write $w=logz$. Now, substituting $w$ by $2iw$ and $z$ by $frac{i−z}{i+z}$, exactly in the same sense we get $2iwin log(frac{i−z}{i+z})$, or, by the usual notation, $2iw=log(frac{i−z}{i+z})$. What was misleading for me is the statement "taking logarithm of both sides". Thanks for the answer.
$endgroup$
– serenus
Jan 20 at 9:34
$begingroup$
What do you mean when saying "some"? Could you please write your answer a bit more explicitly?
$endgroup$
– serenus
Jan 19 at 21:14
$begingroup$
What do you mean when saying "some"? Could you please write your answer a bit more explicitly?
$endgroup$
– serenus
Jan 19 at 21:14
$begingroup$
I've edited my answer. What do you think now?
$endgroup$
– José Carlos Santos
Jan 19 at 21:24
$begingroup$
I've edited my answer. What do you think now?
$endgroup$
– José Carlos Santos
Jan 19 at 21:24
1
1
$begingroup$
I think that $log (e^{2iw})={2iw+i2npi: nin mathbb{Z}}$, and since the sets $log(frac{i-z}{i+z})$ , $log(frac{i-z}{i+z})+i2npi$, are the same, as a consequence we get (1). Am I right?
$endgroup$
– serenus
Jan 19 at 21:42
$begingroup$
I think that $log (e^{2iw})={2iw+i2npi: nin mathbb{Z}}$, and since the sets $log(frac{i-z}{i+z})$ , $log(frac{i-z}{i+z})+i2npi$, are the same, as a consequence we get (1). Am I right?
$endgroup$
– serenus
Jan 19 at 21:42
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Jan 19 at 21:44
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Jan 19 at 21:44
$begingroup$
Although my above statement is true, now I really see what you mean: Given a fixed $zinmathbb{C}$ , the set of solutions for the equation $e^w=z$ is ${ln|z|+i(Argz+2nπ):n inmathbb{Z}}$, which is denoted by $logz$, and although $w$ is a member of the set $logz$, notationally, we write $w=logz$. Now, substituting $w$ by $2iw$ and $z$ by $frac{i−z}{i+z}$, exactly in the same sense we get $2iwin log(frac{i−z}{i+z})$, or, by the usual notation, $2iw=log(frac{i−z}{i+z})$. What was misleading for me is the statement "taking logarithm of both sides". Thanks for the answer.
$endgroup$
– serenus
Jan 20 at 9:34
$begingroup$
Although my above statement is true, now I really see what you mean: Given a fixed $zinmathbb{C}$ , the set of solutions for the equation $e^w=z$ is ${ln|z|+i(Argz+2nπ):n inmathbb{Z}}$, which is denoted by $logz$, and although $w$ is a member of the set $logz$, notationally, we write $w=logz$. Now, substituting $w$ by $2iw$ and $z$ by $frac{i−z}{i+z}$, exactly in the same sense we get $2iwin log(frac{i−z}{i+z})$, or, by the usual notation, $2iw=log(frac{i−z}{i+z})$. What was misleading for me is the statement "taking logarithm of both sides". Thanks for the answer.
$endgroup$
– serenus
Jan 20 at 9:34
|
show 1 more comment
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